A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x,y) is in the relation. 

A function is a type of relation, but a relation is allowed to have the object x in the first set to be related to more than one object in the second set. So, a relation may not be represented by a function machine, because, given the object x to the input of the machine, the machine couldn’t spit out a unique output object that is paired to x.

In mathematics, a binary relation is a general concept that defines some relation between the elements of two sets. It is a generalization of the more commonly understood idea of a mathematical function, but with fewer restrictions. A binary relation over sets X and Y is a set of ordered pairs (x, y) consisting of elements x in X and y in Y. It encodes the common concept of relation: an element x is related to an element y, if and only if the pair (x, y) belongs to the set of ordered pairs that defines the binary relation.

In Maths, the relation is the relationship between two or more set of values. Suppose, x and y are two sets of ordered pairs. And set x has relation with set y, then the values of set x are called domain whereas the values of set y are called range. 

Example: For ordered pairs= {(1,2),(-3,4),(5,6),(-7,8),(9,2)}

How do you write a relation as a set of ordered pairs?

A relation from a set A to a set B is a subset of A×B. Hence, a relation R consists of ordered pairs (a,b), where aA and bB. If (a,b)R, we say that is related to, and we also write aRb.

Question:

 Determine the domain and range of the relation R defined by

R = {x + 2, x + 3} : x {0, 1, 2, 3, 4, 5}

Solution:

Since, x = {0, 1, 2, 3, 4, 5}

Therefore,

      x = 0 x + 2 = 0 + 2 = 2 and x + 3 = 0 + 3 = 3

      x = 1 x + 2 = 1 + 2 = 3 and x + 3 = 1 + 3 = 4

      x = 2 x + 2 = 2 + 2 = 4 and x + 3 = 2 + 3 = 5

      x = 3 x + 2 = 3 + 2 = 5 and x + 3 = 3 + 3 = 6

      x = 4 x + 2 = 4 + 2 = 6 and x + 3 = 4 + 3 = 7

      x = 5 x + 2 = 5 + 2 = 7 and x + 3 = 5 + 3 = 8

Therefore, Domain of R = {a : (a, b) ∈R} = Set of first components of all ordered pair belonging to R.
Therefore, Domain of R = {2, 3, 4, 5, 6, 7}

Range of R = {b : (a, b) ∈ R} = Set of second components of all ordered pairs belonging to R.
Therefore, Range of R = {3, 4, 5, 6, 7, 8}

Important Terms on Relations and Functions

1. Relation R from a non-empty set A to a non-empty set B is a subset of A × B.
2. If n(A) = p, n(B) = q then n (A × B) = pq and number of relations = 2pq
3. A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.
4. Let A and B be two non-empty finite sets such that n(A) = p and n(B) = q then number of functions from A to B = qp
In the set theory, a relation is a way of showing a connection or relationship between any two sets.
Relation
A connection between the elements of two or more sets is Relation. The sets must be non-empty. A subset of the Cartesian product also forms a relation R.

Representation of Relations

Relation defines the relationship between two sets. There are three ways of representing the relations. They are:

  1. Set-Builder Form

The relation can be written in set builder form using the variables x, y in the ordered pair with semi column followed by a relation between x and y.

If A and B are two non-empty sets, then the relation R from A to B is a subset of A×B, i.e., R A × B.

If (a, b) R, then we write a R b and is read as a is related to b.

Question:
The figure shows a relationship between the sets P and Q. Write this relation in
(i) in set-builder form (ii) roster form
Answer:
The relation mentioned in the figure shows, P a domain and Q as range.
Let the relation be R
In roster form R = {(5,3), (6,4), (7,5)}
In set builder form R = {(x,y) : x ∈ P, y ∈ Q, y = x – 2}
[x⇒5,6,7 & y⇒ 3,4,5]
The relation of two sets A={2,3,4} and B={4,9,16}, in which elements of set A are the positive square root of elements of set B. The relation can be written in set builder form as follows.
R=R={(x,y):x is the positive square root of y, x∈A, y∈B }

2. Roaster Form
The relation can be represented in roaster form by writing all the possible ordered pairs of the two sets.
The relation of two sets A={2,3,4}A= and B={4,9,16}, in which elements of set A are the square root of elements of set The B. relation can be written in roaster form as follows.
R={(2,4),(3,9),(4,16)}

3. Arrow Diagram
In this method, the relation between two sets is shown by using the arrow drawn from one set to another set.
The relation of two sets A= {2,3,4} and B= {4,9,16}, in which elements of set A are the square root of elements of set B. The relation can be written by using an arrow diagram below.

Types Of Relations

1. Empty Relation
A relation R in a set A is called empty relation, if no element of A is related to any element of A.
R = φ ⊂ A × A.
We can define void relation as a relation R in a set A, where no element of set A is related to any element of A. So, R = ɸ which is a subset of A × A.

2. Universal Relation
A relation R in a set, say A is a universal relation if each element of A is related to every element of A, i.e., R = A × A. Also called Full relation. Suppose A is a set of all natural numbers and B is a set of all whole numbers. The relation between A and B is universal as every element of A is in set B.

Example : Let A be the set of all students of a boys school. Show that the relation R in A given by R = {(a,b) : difference between the heights of a and b is less than 3 meters} is the universal-relation.
Solution: The difference between the heights of any two students of the school has to be less than 3 meters. Therefore (a,b) ∈ R for all a, b ∈ A.
⇒ R = A X A
⇒ R is the universal-relation on set A.

3. Identity Relation

Every element is related to itself only”

Identity relation graph
If we plat identity function, then it will appear to be a straight
Line. Let us plot a graph function say f(x) = x, by putting different
Values of x.

x

-2 -1 0 1 2
f(x) = y -2 -1 0 1 2

Now, as we can she from above table, the values are the same

For both x-axis and y-axis. Hence, let us plot a graph based

on these values.

Now, from above graph, it is clear that identity function gives a straight 

Line in the xy-plane.

The rule for identity relation is given below.

Let R be a relation defined on the set A,          

If R is identity relation, then

R = {(a, a) / for all a  A}

That is, every element of A has to be related to itself only. 

In case, there is an ordered pair (a, b) in R, then R is not identity. Because the ordered pair (a, b) does not satisfy the above rule of identity relation. 

Examples

Let A = {1, 2, 3} and R be a relation defined on set A as

R = {(1, 1), (2, 2), (3, 3)}

Verify R is identity. 

Solution: 

In the set A, we find three elements. They are 1, 2 and 3.

When we look at the ordered pairs of R, we find the following associations. 

(1, 1) —–> 1 is related to 1

(2, 2) —–> 2 is related to 2

(3, 3) —–> 3 is related to 3

In R, every element of A is related to itself and not to any other different element. 

That is, every element of A is related to itself only. 

So, R is identity.

  • Inverse Relation

If a set with elements has the inverse pairs of another set, then the relation is called inverse relation. An inverse relation is the inverse of a relation and is obtained by interchanging the elements of each ordered pair of the given relation. 

Let R be a relation from a set A to another set B. Then R is of the form {(x, y): x A and y B}. 

The inverse relationship of R is denoted by R-1 and its formula is R-1 = {(y, x): y B and x A}. i.e.,

The first element of each ordered pair of R = the second element of the corresponding ordered pair of R-1 and the second element of each ordered pair of R = the first element of the corresponding ordered pair of R-1.

Inverse relation

Examples on inverse relation:

For, R {(1,2), (4,2), (9,3), (16,4)}, find Domain, Range, and R-1.

Ans. Domain (D) = {1, 4, 9, 16}

Range (R) = {1, 2, 3,4}

R-1 = {(1, 1), (2, 4), (3, 9), 4 ,16)}

  • Reflexive Relation

A relation specified on a set is a reflexive relation if and only if every component of the set is linked to itself. If there is a particular element of the set that is not related to itself, then relation R is not a reflexive relation.

For example, consider a set A = {1,2}. Now, the reflexive relation will be R = {1, 1), (2, 2),

 (I, 2), (2, 1)}

Hence, a relation is reflexive if: (a, a) ∈ R ∀ a ∈ A

[∀ This symbol means for all (or sometimes, for every)

Where a is the element, A is the set and R is the relation.

Example, consider a set A = {p, q, r, s}.vv

The relation R1 = {(p, p), (p, r), (q, q), (r, r), (r, s), (s, s)} in A is reflexive, since every element in A is R1 related to itself.

But the relation R2 = {(p, p), (p, r), (q, r), (q, s), (r, s)} is not reflexive in A since q, r, s A but 

(q, q) ∉ R2, (r, r) ∉ R2 and (s, s) ∉ R2.

  • Symmetric Relation

A relation R on a set A is said to be symmetric if and only if. (a,b) R. and (b,a) R for all a,b A. that is aRb = bRb for all a,b A.

Symmetric relation R defined on a set A for elements a, b  A, we have aRb, that is, (a, b)  R, then we must have bRa, that is, (b, a)  R. This implies that a relation defined on a set A is a symmetric relation if and only if it satisfies aRb bRa for all elements a, b in A. If there is a single ordered pair in R such that (a, b)  R and (b, a) R, then R is not a symmetric relation.

A relation R on a set A is said to be symmetric relation if 

(a, b) R (b, a) R, for all a, b A.

i.e., aRb ⇒ bRa for all a, b ∈ A

Symmetric Relations Examples

Example: 

Suppose R is a relation on a set A where A = {1, 2, 3} and R = {(1,1), (1,2), (1,3), (2,3), (3,1)}. Check if R is a symmetric relation.

Solution: As we can see (1, 2)  R. For R to be symmetric (2, 1) should be in R but (2, 1) R.

Hence, R is not a symmetric relation.

Answer: R = {(1,1), (1,2), (1,3), (2,3), (3,1)} is not a symmetric relation.

Example: 

Suppose R is a relation on a set A where A = {a, b, c} and R = {(a, a), (a, b), (a, c), (b, c), (c, a)}. Determine the elements which should be in R to make R a symmetric relation.

Solution: To make R a symmetric relation, we will check for each element in R.

(a, a)  R (a, a)  R

(a, b)  R (b, a)  R, but (b, a)  R

(a, c)  R (c, a)  R

(b, c)  R (c, b)  R, but (c, b)  R

Hence, (b, a) and (c, b) should belong to R to make R a symmetric relation.

Answer: (b, a) and (c, b) should belong to R to make R a symmetric relation.


  • Transitive Relation

Transitive relations are binary relations defined on a set such that if the first element is related to the second element, and the second element is related to the third element of the set, then the first element must be related to the third element. 

For example, if for three elements a, b, c in set A, if a = b and b = c, then a = c. Here, equality ‘=’ is a transitive relation. 

A relation in a set A is transitive if, (a, b) R, (b, c) R, then (a, c) R, for all a, b, c A

Important Notes on Transitive Relations

  • A relation defined on an empty set is always a transitive relation.
  • There is no fixed formula to determine the number of transitive relations on a set.
  • The complement of a transitive relation need not be transitive.

Transitive Relations Examples:

Example: 

Define a relation R on a set A = {a, b, c} as R = {(a, b), (b, c), (b, b)}. Determine if R is a transitive relation.

Solution: 

As we can see that (a, b) R and (b, c) R, and for R to be transitive (a, c) R must hold, 

but (a, c) R.

So, R is not a transitive relation.

Answer: R is not a transitive relation

 

Example: 

Check if ‘is parallel to’ defined on a set of lines is a transitive relation.

Solution: 

We know that if line 1 is parallel to line 2 and line 2 is parallel to line 3, then line 1 is parallel to line 3 as lines parallel to the same line are parallel to each other.

Answer: ‘Is parallel to’ is a transitive relation.

  • Equivalence Relation

A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive. The equivalence relation is a relationship on the set which is generally represented by the symbol “.

An equivalence relation is a relationship on a set, generally denoted by “”, that is reflexive, symmetric, and transitive for everything in the set. 1. (Reflexivity) a a, 2. (Symmetry) if a b then b a, 3. (Transitivity) if a b and b c then a c.

Relation R is transitive because whenever (a, b) and (b, c) belongs to R, (a, c) also belongs to R. Example: (3, 1) R and (1, 3) R (3, 3) R. So, as R is reflexive, symmetric and transitive, hence, R is an Equivalence Relation.

We say  is an equivalence relation on a set A if it satisfies the following three properties: a ∼ R b”

  1. a) Reflexive: (a, a) ∈ R, for all a∈ A
  2. b) Symmetric: (a, b) ∈ R⇒ (b, a) ∈R 

for all a, b∈ A

  1. c) Transitive: (a, b) ∈R and (b, c) ∈ R

⇒ (a, c) ∈R, for all a, b, c∈ A 

Question:

Let A = {1,2,3,4}. Let R be the equivalence relation on A x A defined by (a, b)R(c, d) iff a + d = b + c.

Find the equivalence class [1, 3].

Solution:

Given A {1,2,3,4}

R is defined as (a, b)R(c, d) iff a + d = b + c

In this relation (a, b) goes in, (c, d) comes out.

We need to find out [(1, 3)]

So, (1, 3) will go in and (c, d) will come out.

This will be possible if a + d = b + c

1 + d = 3 + c (a = 1, b = 3)

d – c = 3 – 1

d – c = 2

So, in our relation [(1, 3)]

We need to find values of c and d which satisfy d – c = 2

Since (c, d) ∈ A x A

Both c and d are in set A = {1,2,3,4}

d – c Numbers (c, d)
2 – 1 = 1 d = 2, c = 1 Not possible
3 – 1 =2 d = 3, c = 1 (1, 3)
3 – 2 = 1 d = 3, c = 2 Not possible
4 – 1 = 3 d = 4, c = 1 Not possible
4 – 2 = 2 d = 4, c = 2 (2, 4)
4 – 3 = 1 d = 4, c = 3 Not possible

 

So, only (1, 3) and (2, 4) satisfy

Hence, {(1, 3)] = {(1, 3), (2, 4)

Question:

 Let m be a positive integer. A relation R is defined on the set Z by “aRb if and only if a – b is divisible by m” for a, b Z. Show that R is an equivalence relation on set Z.

Solution:

(i) Let a Z. Then a – a = 0, which is divisible by m

Therefore, aRa holds for all a Z.

Hence, R is reflexive.

(ii) Let a, b Z and aRb holds. Then a – b is divisible by m and therefore, b – a is also divisible by m.

Thus, aRb bRa.

Hence, R is symmetric.

(iii) Let a, b, c Z and aRb, bRc both hold. Then a – b is divisible by m and b – c is also divisible by m. Therefore, a – c = (a – b) + (b – c) is divisible by m.

Thus,  aRb and bRc aRc

Therefore, R is transitive.

Since, R is reflexive, symmetric and transitive so, R is an equivalence relation on set Z

Functions:

A function in set theory world is simply a mapping of some (or all) elements from Set A to some (or all) elements in Set B. In the example above, the collection of all the possible elements in A is known as the domain; while the elements in A that act as inputs are specially named arguments

The types of functions are defined on the basis of the domain, range, and function expression. The expression used to write the function is the prime defining factor for a function. Along with expression, the relationship between the elements of the domain set and the range set also accounts for the type of function. The classification of functions helps to easily understand and learn the different types of functions.

Representation of Functions

There are three different forms of representation of functions. The functions need to be represented to showcase the domain values and the range values and the relationship between them. The functions can be represented with the help of Venn diagrams, graphical formats, and roster forms. The details of each of the forms of representation are as follows

Venn Diagram: TYPES OF FUNCTONS

The Venn diagram is an important format for representing the function. The Venn diagrams are usually presented as two circles with arrows connecting the element in each of the circles. The domain is presented in one circle and the range values are presented in another circle. And the function defines the arrows, and how the arrows 

connect the different elements in the two circles.

Graphical Form: 

Functions are easy to understand if they are represented in the graphical form with the help of the coordinate axes. Representing the function in graphical form, helps us to understand the changing behaviour of the functions if the function is increasing or decreasing. The domain of the function – the x value is represented along the x-axis, and the range or the f(x) value of the function is plotted with respect to the y-axis.

When Does a Graph Represent a Function?

To say that y is a function of x means that for each value of x there is exactly one value of y. Graphically, this means that each vertical line must intersect the graph at most once. Hence, to determine if a graph represents a function one uses the following simple visual test:

Vertical Line Test:

A graph is a function if and only if every vertical line crosses the graph at most once. According to the vertical line test and the definition of a function, if a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function.

If the vertical line touches the graph at more than one point, then the graph is not a function.

Question:

Represent the function f = {(1, 2), (2,2), (3, 2), (4, 3), (5, 4)} through a graph.

Solution:

F = {(1, 2), (2, 2), (3, 2), (4, 3), (5, 4)}

(1, 2) x =1, f(x) = y = 2

x F(x) = y
1 2
2 2
3 2
4 3
5 4

Roster Form: 

The roster form to represent the set is one of the easiest representations. In roster form, the elements of a set are represented in a row and separated by a comma. For example, the set of first five positive even numbers are represented like: A = {2, 4, 6, 8, 10}.

Example 1: Express the below two sets X and Y in the

roster form.

Solution: The set X in roster form can be expressed like: X = {1, 2, 3, 4}. The set Y in roster form can be expressed like: Y = {D, B, C, A}

Answer: X = {1, 2, 3, 4}, Y = {D, B, C, A}

  • Example 2: Express the set A = {x | x = 2n2 – 1, where n N and n < 5} in roster form.

Solution: The elements of set A are:

  • For n = 1, 2n− 1 = 2 × 12 −1 = 1
  • For n = 2, 2n− 1 = 2 × 2− 1 = 7
  • For n = 3, 2n− 1 = 2 × 3− 1 = 17
  • For n = 4, 2n− 1 = 2 × 4− 1 = 31
  • For n = 5, 2n− 1 = 2 × 5− 1 = 49

Answer: A = {1, 7, 17, 31, 49}

Classification of Function

The types of functions can be broadly classified into four types. 

 1. Based on Element: 

  • One One Function
  • Many One Function
  • Onto Function
  • One One and Onto Function
  • Into Function
  • Constant Function

2. Based on Equation:

  • Identity Function
  • Linear Function
  • Quadratic Function
  • Cubic Function
  • Polynomial Functions

3. Based on the Range:

  • Modulus Function
  • Rational Function
  • Signum Function
  • Even and Odd Functions
  • Periodic Functions
  • Greatest Integer Function
  • Inverse Function
  • Composite Functions

4.  Based on the Domain:

  • Algebraic Functions
  • Trigonometric Functions
  • Logarithmic Functions

Algebraic domains of functions i.e., that is finding the domain of a function

Two big restrictions:

  1. Don’t want Zero in the denominator
  2. (ii) No negative #’s under squire roots

( any when powered root) after simplifying

Example; Find the domain

  1. 1/ (x2 -7x-30 = f(x)     ~
  2. g(x) = √(2x + 3)

BAD, OMIT:

x2 – 7x – 30 = 0

(x – 10) (x + 3) = 0

X =10, x = -3

Domain: All x’s except x = 10, x = -3  

Domain: (-x, -3) U (-3,10) U (10,x)

  1. g(x) = √ (2x + 3)

2x + 3 >=0

2x ≥ -3

X ≥-3/2

Domain: x ≥ -3/2

[-3/2, infinity]

The domain and range of function is the set of all possible inputs and outputs of a function respectively. The domain and range of a function y = f(x) is given as domain= {x ,x R }, range= {f(x), xDomain}. The domain and range of any function can be found algebraically or graphically.

Domain is he set of first co-ordinates. (All possible Input Values)

Range is the set of second co-ordinates. (All possible Output values)

Y = f(x) = x Domain All x F(x) = X2  Domain All x

Range All y Range All y > 0

Two main constraints

Example  f(x) =1/(x-2)

Domain = all x except x = 2 [ DENOMINATOR non zero)

F(x) = (x-3)1/2 

Domain =?

X – 3 ≥ 0 [ equal to or greater than]

X ≥ equal to or greater than 3

(non-negative)1/2  

f(x) = x2 + 1

Range ?

F(x) = 1 as x2 ≥ 0[ equal to or greater than]

One One Function:

A one-to-one function is defined by f: A → B such that every element of set A is connected to a distinct element in set B. The one-to-one function is also called an injective function. Here every A one-to-one function is defined by f: A → B such that every element of set A is connected to a distinct element in set B. The one-to-one function is also called an injective function. Here every element of the domain has a distinct image or co-domain element for the given function.

Many to One Function:

A many to one function is defined by the function f: A → B, such that more than one element of the set A are connected to the same element in the set B. In a many to one function, more than one element has the same co-domain or image. If a many to one function, in the codomain, is a single value or the domain element are all connected to a single element, then it is called a constant function.

  • Onto Function

In an, onto function, every codomain element is related to the domain element. For a function defined by f: A → B, such that every element in set B has a pre-image in set A. The onto function is also called a subjective function.

  • One One and Onto Function (Bijection)

A function that is both a one and onto function is called a bijective function. Here every element of the domain is connected to a distinct element in the codomain and every element of the codomain has a pre-image. Also, in other words every element of set A is connected to a distinct element in set B, and there is not a single element in set B which has been left out.

  • Into Function

Into function is exactly opposite in properties to an onto function. Here there are certain elements in the co-domain that do not have any pre-image. The elements in set B are excess and are not connected to any elements in set A.

  • Constant Function

A constant function is an important form of a many to one function. In a constant function, all the domain elements have a single image. The constant function is of the form f(x) = K, where K is a real number. For the different values of the domain(x value), the same range value of K is obtained for a constant function.


  • Based on Equation:

The algebraic expressions are also functions and are based on the degree of the polynomial. The functions based on equations are classified into the following equations based on the degree of the variable ‘x’.


  • Identity Function

The identity function has the same domain and range. The identity function equation is f(x) = x, or y = x. The domain and range of the identity function is of the form {(1, 1), (2, 2), (3, 3), (4, 4) …. (n, n)}.

The graph of the identity function is a straight line that is equally inclined to the coordinate axes and is passing through the origin. The identity function can take both positive and negative values and hence it is present in the first and the third quadrants of the coordinate axis.

  • Linear Function

A polynomial function having the first-degree equation is a linear function. The domain and range of a linear function is a real number, and it has a straight-line graph. Equations such as y = x + 2, y = 3x, y = 2x – 1, are all examples of linear functions. The identity function of y = x can also be considered a linear function.

Example:

Let R be the set of real numbers. Define the

Real function f: R R by f(x) = x + 10 and sketch

The graph of this function.

Solution:

Here, f(0) = 10, f(1) = 11, f(2) = 12’….

f(10) = 20, etc., and f(-1) = 9, f(-2) = 8 ….f(-10) = 0 and so on.

Therefore, shape of the graph of the function assumes the form as shown here. Hence, the given function is an example of a linear function.

Example:

Let R be the set of real numbers. Define the

Real function f: R R by f(x) = x + 10 and sketch

The graph of this function.

Solution:

Here, f(0) = 10, f(1) = 11, f(2) = 12’….

f(10) = 20, etc., and f(-1) = 9, f(-2) = 8 ….f(-10) = 0 and so on.

Therefore, shape of the graph of the function assumes the form as shown here. Hence, the given function is an example of a linear function.

  • Quadratic Function

A quadratic function has a second-degree quadratic equation and it has a graph in the form of a curve. The general form of the quadratic function is f(x) = ax2 + bx + c, where a ≠ 0 and a, b, c are constants and x is a variable. The domain and range of the quadratic function is R.

The graph of a quadratic equation is a non-linear graph and is parabolic in shape. Examples of quadratic functions are f(x) = 3x2 + 5, f(x) = x2 – 3x + 2.

  • Cubic Function

A cubic function has an equation of degree three. The general form of a cubic function is f(x) = ax3 + bx2 + cx +d, where a ≠ 0 and a, b, c, and d are real numbers & x is a variable. The domain and range of a cubic function is R.

The graph of a cubic function is more curved than the quadratic function. An example of cubic function is f(x) = 8x3 + 5x2 + 3.

  • Polynomial Function

The general form of a polynomial function is f(x) = an xn + an-1 xn-1 + an-2 xn-2+ …. ax + b. Here n is a nonnegative integer and x is a variable. The domain and range of a polynomial function are R. Based on the power of the polynomial function, the functions can be classified as a quadratic function, cubic function, etc.

  • Based on Range:

Here the types of functions have been classified based on the range which is obtained from the given functions. The different types of functions based on the range are as follows.

  • Modulus Function

The modulus function gives the absolute value of the function, irrespective of the sign of the input domain value. The modulus function is represented as f(x) = |x|. The input value of ‘x’ can be a positive or a negative expression

If f(x) is a real number, we define

|x| = x if x ≥ 0

|x| = -x if x ≤ 0

|x| is called the modulus, or the absolute value of x.

For example, |2| = 2, |-5| = 5, |0| = 0.

The function f : R→R defined by f(x) = |x| is known as the modulus function.

The graph of a modulus function lies in the first and the second quadrants since the coordinates of the points on the graph are of the form (x, y), (-x, y).

  1. Rational Function

A function that is composed of two functions and expressed in the form of a fraction is a rational function. A rational fraction is of the form f(x)/g(x), and g(x) ≠ 0. The functions used in this rational function can be an algebraic function or any other function. The graphical representation of these rational functions is similar to the asymptotes, since it does not touch the axis lines.

  • Signum Function

This function is given by 

Now plotting graph:

        x< 0         x= 0       x > 0
x -1 -2 0 0 1 2
y -1 -1 0 O 1 1

Here, Domain = All values of x = R

Range = All values of y

Since, y will have value 0, 1 or -1

Range = {0, 1, -1}

The signum function helps us to know the sign of the function and does not give the numeric value or any other values for the range. The range of the signum function is limited to {-1, 0, 1}. For the positive value of the domain, the signum function gives an answer of 1, for negative values the signum function gives an answer of -1, and for the 0 value of a domain, the image is 0. The signum function has wide application in software programming.

  • Even and Odd Function

The even and odd functions are based on the relationship between the input and the output values of the function. For the negative domain value, if the range is a negative value of the range of the original function, then the function is an odd function. And for the negative domain value, if the range is the same as that of the original function, then the function is an even function.

A function of x if f(- x) = – f(x) and is an even function of x if f(-x) = f(x).

Example: Let f(x) = x. Now, f(-x) = -x = – f(x)

So, f(x) = x is an odd function of x.

For f(x) =x2 , f(-x) = (-x)2 = x2 ;

So, f(x) = x2 is an even function of x.

An example of even functions are x2, Cosx, Secx, and an example of odd functions are x3, Sinx, Tanx.

f(x) = x2 is an even function of x

  • Periodic Function

The function is considered a periodic function if the same range appears for different domain values and in a sequential manner. The trigonometric functions can be considered periodic functions. For example, the function f(x) = sin x, have a range [-1, 1] for the different domain values of

 x = nπ + (-1)n x. Similarly, we can write the domain and the range of the trigonometric functions and prove that the range shows up in a periodic manner.

  • Inverse function

The inverse of a function f(x) is denoted by f-1(x). For inverse of a function the domain and range of the given function is changed as the range and domain of the inverse function. The inverse of a function exists, if it is a bijective function.

If a function f(x) = x2, then the inverse of the function is f-1(x) = √x.

  • Greatest Integer Function

The function f: R ⟶R defined by f(x) =[x]

x ∈ R assumes the value of the greatest integer, less than or equal to x. Such a function is called the greatest integer function.

From the definition of[x], we can see that

[x] = -1 for -1 x < 0

[x] = 0 for 0 ≤ x < 1

[x] = 1 for 1 ≤ x < 2

[x] = 2 for 2 ≤ x < 3 and so on

The graph of the equation is shown in the figure. f(x) = [x]

The greatest integer function graph is known as the step curve because of the step structure of the 

curve.

  • Composite Function

The composite functions are of the form of gof(x), fog(x), h(g(f(x))), and is made from the individual functions of f(x), g(x), h(x). The composite functions made of two functions have the range of one function forming the domain for another function. Let us consider a composite function fog(x), which is made up of two functions f(x) and g(x).

Here we write fog(x) = f(g(x)). The range of g(x) forms the domain for the function f(x). It can be considered as a sequence of two functions. If f(x) = 2x + 3 and g(x) = x + 1 we have fog(x) = f(g(x)) = f(x + 1) = 2(x + 1) + 3 = 2x + 5.

  • Functions – Based on Domain:

The three broad types of functions based on domain value are as follows.

  • Algebraic Function

An algebraic function is helpful to define the various operations of algebra. The algebraic function has a variable, coefficient, constant term, and various arithmetic operators such as addition,

subtraction, multiplication, division. An algebraic function is generally of the form of f(x) = anxn + an – 1xn – 1+ an-2xn-2+ ……. ax + c.

The algebraic function can also be represented graphically. The algebraic function is also termed as a linear function, quadratic function, cubic function, polynomial function, based on the degree of the algebraic equation.

  • Trigonometric Functions

The trigonometric functions also have a domain and range similar to any other function. The six trigonometric functions are f(θ) = sinθ, f(θ) = cosθ, f(θ) = tanθ, f(θ) = secθ, f(θ) = cosecθ. Here the domain value θ is the angle and is in degrees or in radians. These trigonometric functions have been taken based on the ratio of the sides of a right-angle triangle, and are based on the Pythagoras theorem.

Further from these trigonometric functions, inverse trigonometric functions have also been derived. The domain of the inverse trigonometric function is a real number value and its range is an angle. The trigonometric functions and the inverse trigonometric functions are also sometimes referred to as periodic functions since the principal values are repeated.

  • Logarithmic Functions

Logarithmic functions have been derived from the exponential functions. The logarithmic functions are considered as the inverse of exponential functions. Logarithmic functions have a ‘log’ in the function and it has a base. The logarithmic function is of the form y = logb x

A logarithmic function is a function of the form. which is read “y equals the log of x, base b” or “y” equals the log, base b, of x.” In both forms, x > 0 and b > 0, b ≠ 1. There are no restrictions on y.

Here the domain value is the input value of ‘x’ and is calculated using the Napier logarithmic table. The logarithmic function gives the number of exponential times to which the base has raised to obtain the value of x. The same logarithmic function can be expressed as an exponential function as x = by

Question 1: The function f: [0, 3] -> [1, 29] defined by f(x) = 2x3 – 15x2 + 36x + 1 is

(a) one-one and onto

(b) onto but not one-one

(c) one-one but not onto

(d) neither one-one nor onto

Solution:

f(x) = 2x3 – 15x2 + 36x + 1

f(x) = 6x2 – 30x + 36

       = -6(x2 – 5x + 6)

       = 6 (x – 2) (x – 3)

F(x) is increasing in [0, 2] and decreasing in [2, 3]. 

We know, if function is strictly increasing or decreasing in its domain, then it is one-one. But given function is increasing as well as decreasing.

So, f(x) is many one.

f(0) = 1

f(2) = 29

f(3) = 28

Range is [1, 29]

Hence, f(x) is onto.

Answer: Correct option (B) onto but not one-one

Question 2: Let f(x) = x2 and g(x) = sin x for all x Є R. Then the set of all x satisfying (f o g o g o f)(x) = (g o g o f)(x), where (f o g)(x) = f(g(x)) is.

  1. ± n, n Є {0,1, 2…}
  2. ± , n Є {1, 2…}
  3.  2 + 2n, n Є {…, -2, -1, 0,1, 2…}
  4.  2n, n Є {…, -2, -1, 0,1, 2…}